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Quantum Mechanics - Ladder Operators

  • Thread starter Tangent87
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  • #1
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I'm trying to show that [tex]\sum_{m=0}^\infty \frac{1}{m!} (-1)^m {a^{\dagger}}^m a^m =|0 \rangle\left\langle 0|[/tex]

Where a and [tex]{a^{\dagger}}[/tex] denote the usual annihilation and creation operators. The questions suggests acting both sides with |n> but even if I did that and showed LHS=...=RHS then that still doesn't prove the original expression (we can't reverse the implies sign if you see what I mean). So I'm stuck as to what to do.
 

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  • #2
dextercioby
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You have an equality of operators. The LHS is a linear operator, the RHS is a linear operator as well. On a vector space, two operators are equal iff their domains and codomains are equal. The projection operator on the RHS clearly is defined on all the Hilbert (Fock) space of the problem (as it is bounded), so my guess is that, if you can't show that the LHS is also bounded and defined on all H/F space, at least you could assume that the operatorial relation holds on the domain of the operator in the LHS (which would be equal to the common domain).

So you're only supposed to show that the codomains are equal, which you claim to have done, right ? If not, then write the sum explicitely and use how the operators act on an arbitrary state |n>.

Then you're done, I guess.
 
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You have an equality of operators. The LHS is a linear operator, the RHS is a linear operator as well. On a vector space, two operators are equal iff their domains and codomains are equal. The projection operator on the RHS clearly is defined on all the Hilbert (Fock) space of the problem (as it is bounded), so my guess is that, if you can't show that the LHS is also bounded and defined on all H/F space, at least you could assume that the operatorial relation holds on the domain of the operator in the LHS (which would be equal to the common domain).

So you're only supposed to show that the codomains are equal, which you claim to have done, right ? If not, then write the sum explicitely and use how the operators act on an arbitrary state |n>.

Then you're done, I guess.
Hmm ok, I'll be honest I didn't really understand most of what you said but if you say all I need to do is verify it for one case then I'll trust you and leave it at that. Thanks for your help.
 
  • #4
dextercioby
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On a second thought, you can try to <sandwich> the operatorial equality between <n| and |n>. What do you get ?
 
  • #5
fzero
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The questions suggests acting both sides with |n> but even if I did that and showed LHS=...=RHS then that still doesn't prove the original expression (we can't reverse the implies sign if you see what I mean).
This statement is incorrect. If you're given two operators [tex]\hat{A},\hat{B}[/tex] and you can show that

[tex] (\hat{A} -\hat{B}) |n\rangle =0, ~\forall n,[/tex]

where [tex]|n\rangle[/tex] is a complete set of states, then we can conclude that the operators are equal, [tex]\hat{A}=\hat{B}[/tex].
 

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